Question

Prove that the points A(−3, 0), B(1, −3) and C(4, 1) are the vertices of an isosceles right-angled triangle. Find the area of this triangle.

Answer 1

Let $A\left(-3,0\right),B\left(1,-3\right)\text{and}C\left(4,1\right)$ be the given vertices. Then,
$$\begin{gathered} AB=\sqrt{{\left(1-\left(-3\right)\right)}^2+{\left(-3-0\right)}^2} \\ =\sqrt{{\left(4\right)}^2+{\left(-3\right)}^2} \\ =\sqrt{16+9} \\ =\sqrt{25} \\ =5\text{units} \\ BC=\sqrt{{\left(4-1\right)}^2+{\left(1-\left(-3\right)\right)}^2} \\ =\sqrt{{\left(3\right)}^2+{\left(4\right)}^2} \\ =\sqrt{9+16} \\ =\sqrt{25} \\ =5\text{units} \\ AC=\sqrt{{\left(4-\left(-3\right)\right)}^2+{\left(1-0\right)}^2} \\ =\sqrt{{\left(7\right)}^2+{\left(1\right)}^2} \\ =\sqrt{49+1} \\ =\sqrt{50} \\ =5\sqrt{2}\text{units} \end{gathered}$$
Thus, AB = BC = 5 units.
Therefore, $∆ABC$ is an isosceles triangle.
Also,
$\left(A{B}^2+B{C}^2\right)=\left(5^2+5^2\right)$
= 50 units.
$\text{and}A{C}^2={\left(5\sqrt{2}\right)}^2$
= 50 units
Thus, $A{B}^2+B{C}^2=A{C^2}^{}$
This shows that $∆ABC$ is right-angled at B.
So, ∆ABC is an isosceles right-angled triangle.
In $∆ABC$, we have:
base BC = 5 units and height AB= 5 units
$\text{Area of}∆ABC=\left(\frac{1}{2}\times \text{base}\times \text{height}\right)$
$$\begin{gathered} =\left(\frac{1}{2}\times 5\times 5\right) \\ =12.5\text{sq units} \end{gathered}$$